a ball is gently dropped from a height of 20m.if its velocity increases uniformly at the rate of 10m/s^2.with what velocity will it strike the ground and after what time will it strike the ground?
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4
given,
h=20m,g=10m/s2 ,u=0
t=sq root of 2h/g
=sq root of 2×20/10=sq root of 4=2 sec
t=2sec
V=u+gt=10×2=20m/s
h=20m,g=10m/s2 ,u=0
t=sq root of 2h/g
=sq root of 2×20/10=sq root of 4=2 sec
t=2sec
V=u+gt=10×2=20m/s
balu135:
u+gt means
v=u+at it is one equations of motion
U+at
Answered by
5
ɢɪᴠᴇɴ
ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ(ᴜ) = 0 ᴍ/ꜱ
ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ (ᴀ) = 10 ᴍ/ꜱ²
ᴛɪᴍᴇ(ꜱ) = 20 ᴍ
ᴡᴇ ᴋɴᴏᴡ
ᴠ² = ᴜ² + 2ᴀꜱ
ᴘᴜᴛᴛɪɴɢ ᴠᴀʟᴜᴇꜱ:
ᴠ² = 0² + 2 (10 x 20)
ᴠ² = 400
ᴠ = 20 ᴍ/ꜱ
ꜰᴏʀ ᴛɪᴍᴇ:
ᴠ = ᴜ + ᴀᴛ
ᴛ = ᴠ - ᴜ/ᴀ
ᴛ = (20-0)10
= 20/10
= 2
∴ ꜱᴛʀɪᴋɪɴɢ ᴠᴇʟᴏᴄɪᴛʏ = 20 ᴍ/ꜱ
∴ ꜱᴛʀɪᴋɪɴɢ ᴛɪᴍᴇ = 2 ꜱᴇᴄᴏɴᴅꜱ
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